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  "body": "\n## Definition\n\nLet $K$ be a [[commutative ring|0GYGHKX834]].\nLet $A$ be a [[ring|0894J980RK]] whose additive group carries a [[$K$-module|09VDS2W1E8]] structure.\nThen $A$ is an *associative algebra over $K$*, or *$K$-algebra*, if\n$$k(ab) = (ka)b = a(kb)$$\nfor all $k \\in K$ and $a, b \\in A$.\n\n## Remarks\n\nEquivalently, a $K$-algebra is a ring $A$ with a ring homomorphism $\\eta \\colon K \\to A$ whose image lies in the center of $A$; the scalar product is then $ka = \\eta(k)a$.\nExample: for a field $F$, the matrix ring $M_n(F)$ is an $F$-algebra with the entrywise scalar product.\nThe check: $k(AB)$, $(kA)B$ and $A(kB)$ all have $(i,j)$ entry $k \\sum_l a_{il} b_{lj}$, because $k$ commutes with the entries.\nNon-example: the quaternions $\\mathbb{H}$ with left multiplication by elements of $\\C \\subset \\mathbb{H}$ are not a $\\C$-algebra.\nThe condition $k(ab) = a(kb)$ fails for the scalar $i$ with $a = j$ and $b = 1$, since $i(j \\cdot 1) = ij$ while $j(i \\cdot 1) = ji = -ij \\ne ij$.\nThey are an algebra over $\\R$, because real numbers commute with all quaternions.\nEvery ring is a $\\Z$-algebra in exactly one way.\nSome authors do not require a unit element in $A$; this note assumes one.\n",
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